Holt Algebra 1

2-Ext Solving Absolute-Value Equations2-Ext Solving Absolute-Value Equations

Holt Algebra 1

Lesson PresentationLesson Presentation

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Solve equations in one variable that contain absolute-value expressions.

Objective

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

The absolute-value of a number is that numbers distance from zero on a number line. For example, |–5| = 5.

5 4 3 2 0 1 2 3 4 56 1 6

5 units

Both 5 and –5 are a distance of 5 units from 0, so both 5 and –5 have an absolute value of 5.

To write this using algebra, you would write |x| = 5. This equation asks, “What values of x have an absolute value of 5?” The solutions are 5 and –5. Notice this equation has two solutions.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

To solve absolute-value equations, perform inverse operations to isolate the absolute-value expression on one side of the equation. Then you must consider two cases.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Example 1A: Solving Absolute-Value Equations

Solve each equation. Check your answer.

|x| = 12|x| = 12

Case 1 x = 12

Case 2 x = –12

The solutions are 12 and –12.

Think: What numbers are 12 units from 0?

Rewrite the equation as two cases.

Check |x| = 12

|12| 1212 12

|x| = 12

|12| 1212 12

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Example 1B: Solving Absolute-Value Equations

3|x + 7| = 24

|x + 7| = 8

The solutions are 1 and –15.

Case 1 x + 7 = 8

Case 2 x + 7 = –8

– 7 –7 – 7 – 7x = 1 x = –15

Since |x + 7| is multiplied by 3, divide both sides by 3 to undo the multiplication.

Think: What numbers are 8 units from 0?

Rewrite the equations as two cases. Since 7 is added to x subtract 7 from both sides of each equation.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Example 1B Continued

3|x + 7| = 24

The solutions are 1 and –15.

Check 3|x + 7| = 24 3|x + 7| = 24

3|8| 24

24 24

3|1 + 7| 24

3(8) 24

3|15 + 7| 24

24 24

3|8| 24

3(8) 24

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Solve each equation. Check your answer.

Check It Out! Example 1a

|x| – 3 = 4|x| – 3 = 4

+ 3 +3|x| = 7

Case 1 x = 7

Case 2 –x = 7

–1(–x) = –1(7)x = –7x = 7

The solutions are 7 and –7.

Since 3 is subtracted from |x|, add 3 to both sides.

Think: what numbers are 7 units from 0?

Rewrite the case 2 equation by multiplying by –1 to change the minus x to a positive..

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Check |x| 3 = 4 |x| 3 = 4

7 3 4

|7| 3 4

4 4

| 7| 3 4

7 3 4

4 4

Solve the equation. Check your answer.

Check It Out! Example 1a Continued

|x| 3 = 4

The solutions are 7 and 7.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Solve the equation. Check your answer.

Check It Out! Example 1b

|x 2| = 8

|x 2| = 8 Think: what numbers are 8 units from 0?

+2 +2

Case 1 x 2 = 8

x = 10+2 +2x = 6

Case 2 x 2 = 8

Rewrite the equations as two cases. Since 2 is subtracted from x add 2 to both sides of each equation.

The solutions are 10 and 6.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Solve the equation. Check your answer.

Check It Out! Example 1b Continued

|x 2| = 8

The solutions are 10 and 6.

Check |x 2| = 8 |x 2| = 8

10 2| 8

|10 2| 8

8 8

| 6 + (2)| 8

6 + 2 8

8 8

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Not all absolute-value equations have two solutions. If the absolute-value expression equals 0, there is one solution. If an equation states that an absolute-value is negative, there are no solutions.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Example 2A: Special Cases of Absolute-Value Equations

Solve the equation. Check your answer.8 = |x + 2| 8

8 = |x + 2| 8+8 + 8

0 = |x +2|

0 = x + 22 22 = x

Since 8 is subtracted from |x + 2|, add 8 to both sides to undo the subtraction.

There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Example 2A Continued

Solve the equation. Check your answer.8 = |x +2| 8

8 8

8 |2 + 2| 8

8 |0| 8

8 0 8

To check your solution, substitute 2 for x in your original equation.

Solution is x = 2

Check 8 =|x + 2| 8

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Example 2B: Special Cases of Absolute-Value Equations

Solve the equation. Check your answer.

3 + |x + 4| = 0

3 + |x + 4| = 03 3

|x + 4| = 3

Since 3 is added to |x + 4|, subtract 3 from both sides to undo the addition.

Absolute values cannot be negative.

This equation has no solution.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Remember!

Absolute value must be nonnegative because it represents distance.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Check It Out! Example 2a

Solve the equation. Check your answer.

2 |2x 5| = 7

2 |2x 5| = 72 2

|2x 5| = 5

Since 2 is added to |2x 5|, subtract 2 from both sides to undo the addition.

Absolute values cannot be negative.

|2x 5| = 5

This equation has no solution.

Since |2x 5| is multiplied by a negative 1, divide both sides by negative 1.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Check It Out! Example 2b

Solve the equation. Check your answer.

6 + |x 4| = 6

6 + |x 4| = 6+6 +6

|x 4| = 0

x 4 = 0+ 4 +4

x = 4

Since 6 is subtracted from |x 4|, add 6 to both sides to undo the subtraction.

There is only one case. Since 4 is subtracted from x, add 4 to both sides to undo the addition.

Holt Algebra 1

2-Ext Solving Absolute-Value Equations

Solve the equation. Check your answer.

Check It Out! Example 2b Continued

6 + |x 4| = 6

6 + |4 4| 6

6 +|0| 6

6 + 0 6

6 6

6 + |x 4| = 6

The solution is x = 4.

To check your solution, substitute 4 for x in your original equation.